Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{x - 2}{x^2 - 5x + 6} \div \dfrac{-2x + 16}{-7x + 21} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{x - 2}{x^2 - 5x + 6} \times \dfrac{-7x + 21}{-2x + 16} $ First factor the quadratic. $k = \dfrac{x - 2}{(x - 3)(x - 2)} \times \dfrac{-7x + 21}{-2x + 16} $ Then factor out any other terms. $k = \dfrac{x - 2}{(x - 3)(x - 2)} \times \dfrac{-7(x - 3)}{-2(x - 8)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (x - 2) \times -7(x - 3) } { (x - 3)(x - 2) \times -2(x - 8) } $ $k = \dfrac{ -7(x - 2)(x - 3)}{ -2(x - 3)(x - 2)(x - 8)} $ Notice that $(x - 2)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -7(x - 2)\cancel{(x - 3)}}{ -2\cancel{(x - 3)}(x - 2)(x - 8)} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $k = \dfrac{ -7\cancel{(x - 2)}\cancel{(x - 3)}}{ -2\cancel{(x - 3)}\cancel{(x - 2)}(x - 8)} $ We are dividing by $x - 2$ , so $x - 2 \neq 0$ Therefore, $x \neq 2$ $k = \dfrac{-7}{-2(x - 8)} $ $k = \dfrac{7}{2(x - 8)} ; \space x \neq 3 ; \space x \neq 2 $